# Addition Exponents

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06 October 06:17

Exponents or admiral are a

Exponents are commonly accounting in the anatomy $a^b$, area $a$ is the abject and $b$ is the exponent. In contexts area superscripts are not available, such as in some contexts in computers, $a^b$ is frequently accounting as a^b or beneath generally as a

When the backer is a absolute integer, then it is just a simple case of adding the abject by itself a assertive amount of

$3^4\; =\; 3$

Here, 3 is the base, 4 is the backer (written as a superscript), and 81 is 3 aloft to the 4th power.

Notice that the abject 3 appears 4 times in the again multiplication, because the backer is 4.

Some added examples:

$$

egin

12^2=12 imes12=1442^8=2 imes2 imes2 imes2 imes2 imes2 imes2 imes2=2561^5=1 imes1 imes1 imes1 imes1=1

end

If you accept two or added exponents with the aforementioned base, then adding them has the aforementioned aftereffect as abacus their exponents.

For instance $(a^b)$

$(3^4)$

If you accept two or added exponents with the aforementioned base, then adding them has the aforementioned aftereffect as adding their exponents.

For instance $(a^b)/(a^c),$ is the aforementioned as $a^,$. For example,

$(3^6)/(3^2)=(3$

# What are $4^3,$, $3^4,$, $1^,$, $^1,$

# Address these numbers as admiral of 2: $128,\; 8,\; 1024,$

# What is $(2^3)\#\; What\; is$ (2^6)/(2^2),$$

# Harder: Why does $3^0=1,$ (clue: anticipate about $/,$, for example)

Click actuality for .

Negative exponents plan hardly differently. Lets say you wish to account $3^$. To do that, you yield $1/3^2$ to get your answer. We do the backer first, see

$3^=frac=frac$

The capricious acreage doesnt administer in exponents. See for yourself! Try to account 2

Exponents do, however, accept their own set of axioms that they consistantly follow. Constant with the above-mentioned examples, one can accompaniment about that:

:$$

(a^b) imes(a^c)=a^

,,,, and

:$$

(a^b)/(a^c)=a^

,,,,

Its aswell simple to see that

$$

egin

(a^b)^c=a^

end

So far, we accept alone apparent exponents as accomplished numbers, but exponents can be apportioned as well. With a apportioned exponent, the numerator acts as a accustomed whole-number exponent, while the denominator acts as a root.

In general,

$a^=sqrt[q],,,,$

for any absolute amount $q$ ≠ 0.

Lets attending at $8^$ as an example. First, we accession 8 to the ability of the numerator, 2. Then, back the denominator is 3, we yield the third basis of this number. The announcement is apprehend as the cubed basis of eight squared, and accounting as:

$8^=sqrt[3]=sqrt[3]=4$

It should then be axiomatic that if the numerator of the apportioned backer is 1, the announcement is a simple root. That is, $1/2$ is a aboveboard root, $1/3$ is a cubed root, $1/4$ is a fourth root, etc.

For example, $9^$ would be apprehend as the aboveboard basis of nine, and accounting as:

$9^=sqrt[2]=sqrt=3$

See also:

Exponents or admiral are a

**action**of again multiplication, in abundant the aforementioned way that multiplication is a action of again addition.Exponents are commonly accounting in the anatomy $a^b$, area $a$ is the abject and $b$ is the exponent. In contexts area superscripts are not available, such as in some contexts in computers, $a^b$ is frequently accounting as a^b or beneath generally as a

When the backer is a absolute integer, then it is just a simple case of adding the abject by itself a assertive amount of

**times**. For example,Here, 3 is the base, 4 is the backer (written as a superscript), and 81 is 3 aloft to the 4th power.

Notice that the abject 3 appears 4 times in the again multiplication, because the backer is 4.

Some added examples:

egin

12^2=12 imes12=1442^8=2 imes2 imes2 imes2 imes2 imes2 imes2 imes2=2561^5=1 imes1 imes1 imes1 imes1=1

end

If you accept two or added exponents with the aforementioned base, then adding them has the aforementioned aftereffect as abacus their exponents.

For instance $(a^b)$

If you accept two or added exponents with the aforementioned base, then adding them has the aforementioned aftereffect as adding their exponents.

For instance $(a^b)/(a^c),$ is the aforementioned as $a^,$. For example,

# What are $4^3,$, $3^4,$, $1^,$, $^1,$

# Address these numbers as admiral of 2: $128,\; 8,\; 1024,$

# What is $(2^3)\#\; What\; is$ (2^6)/(2^2),$$

# Harder: Why does $3^0=1,$ (clue: anticipate about $/,$, for example)

Click actuality for .

Negative exponents plan hardly differently. Lets say you wish to account $3^$. To do that, you yield $1/3^2$ to get your answer. We do the backer first, see

The capricious acreage doesnt administer in exponents. See for yourself! Try to account 2

^{3}, and then see if it is the aforementioned as 3

^{2}(). The distributive and akin backdrop dont administer either.

Exponents do, however, accept their own set of axioms that they consistantly follow. Constant with the above-mentioned examples, one can accompaniment about that:

:$$

(a^b) imes(a^c)=a^

,,,, and

:$$

(a^b)/(a^c)=a^

,,,,

Its aswell simple to see that

$$

egin

(a^b)^c=a^

end

So far, we accept alone apparent exponents as accomplished numbers, but exponents can be apportioned as well. With a apportioned exponent, the numerator acts as a accustomed whole-number exponent, while the denominator acts as a root.

In general,

$a^=sqrt[q],,,,$

for any absolute amount $q$ ≠ 0.

Lets attending at $8^$ as an example. First, we accession 8 to the ability of the numerator, 2. Then, back the denominator is 3, we yield the third basis of this number. The announcement is apprehend as the cubed basis of eight squared, and accounting as:

It should then be axiomatic that if the numerator of the apportioned backer is 1, the announcement is a simple root. That is, $1/2$ is a aboveboard root, $1/3$ is a cubed root, $1/4$ is a fourth root, etc.

For example, $9^$ would be apprehend as the aboveboard basis of nine, and accounting as:

See also:

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Article In : Reference & Education - Mathematics